Can the Principle of Constant Light Speed be Proved by the MMX?

[vuquta]

http://www.lmgtfy.com/?q=aether+wind

 

OMG, you are talking to me.

 

Is this true?

[Xinwei Huang]

Where do you get 0.02%? Sagnac for the earth is just over 200 ns, for a travel time of about 0.13 sec, which is 1.5 parts in 10^6.

 

Hi swansont,

I think you do not understand what I say.

Please note, not the Earth, but the planet.

 

The link pic1.xilu.com/1/3118/6237719/d6c698c33153555f2d41488b9769ec82.jpg is also failed. I do not know why.

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What is a ether wind?

How does it have a speed?

What is it?

Since MMX has null results, how do you compare your logic to MMX?

Are you claiming MMX is wrong or are you claiming the two directions are wrong.

I am just not seeing a logical argument.

 

Hi vuquta,

Yesterday I said, if you think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%, so the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal.

Do you think so?

[vuquta]

Hi swansont,

I think you do not understand what I say.

Please note, not the Earth, but the planet.

 

The link pic1.xilu.com/1/3118/6237719/d6c698c33153555f2d41488b9769ec82.jpg is also failed. I do not know why.

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Consecutive posts merged

 

 

Hi vuquta,

Yesterday I said, if you think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%, so the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal.

Do you think so?

 

I see in ther forums, they think the light will return at the same time.

 

I am in agreement with this forum that Sagnac would dictate an unequal travel.

 

Then you want to cut the two paths into small pieces and then claim these small pieces I suppose put end to end would be different.

 

But that is not what MMX demonstrates and you did mention MMX.

 

See, I am just not seeing how you are tying these two together. Then you bring up an aether wind and have not precisely defined it. In order to invent an object, you would need experimental data evidence to support this new object. Thoughts and math theory would not be sufficient in my mind. This in my view is why your paper was rejected. You invented an object your cannot verify.

 

Finally, what is your math for the .02% timing differential?

[Xinwei Huang]

what is your math for the .02% timing differential?

It looks like that you have not understood what I was saying.

My paper has said that the maximum linear velocity of rotation on the hypothetical star's surface is equal to the linear velocity of Earth revolution, about 30 km•s-1.

Can't you count it?

Analysis from the inertial system without rotation, they can not return to the starting point at the same time.

But most people have not realized that analysis from the planet surface is correct.

Only a few people agree with the latter.

Although my paper has explained why, but most people still do not understand.

This is a very difficult task. I do not know how to do to make most people aware their wrong.

[swansont]

Analysis from the inertial system without rotation, they can not return to the starting point at the same time.

 

But how much phase accumulates? That is the question.

[Xinwei Huang]

But how much phase accumulates? That is the question.

 

This question is very simple.

t1=2πR/(C-V)

t2=2πR/(C+V)

t1/t2=(C+V)/(C-V)=(300000-30)/(30000+30)≈0.9998

This is why the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%.

[swansont]

That's the amount of phase that accumulates due to the rotation. That's not how much phase accumulates due to the revolution about the sun.

[Xinwei Huang]

That's the amount of phase that accumulates due to the rotation. That's not how much phase accumulates due to the revolution about the sun.

 

It seems that you still do not understand my question.

That's the amount of phase that accumulates due to the revolution about the sun.

Please note that 300,000 is the speed of light and 30 is revolution speed of the Earth around the sun.

[swansont]

Then you are doing the calculation incorrectly. Physically, why is the path a different length for clockwise vs counterclockwise travel? When the object is rotating, it's because the velocity vector of the light travel is parallel or antiparallel to the light travel — that's why you have c+v and c-v; it applies for the entire path. But for the effect of revolution this isn't true.

[Xinwei Huang]

Many people can not understand why two beams of light can return to point O at the same time.

Michelson-Morley experiment let us know that the speeds of light in each direction are equal.

The question now is whether their journey is equal after gyrating a circuit they return to point O.

Many people think that they are not equal. Because of the rotation of the planet, the optical path of the light in counter-clockwise is less than that of the light in clockwise direction.

Although I remind that they analyzes this issue in the inertial system that never rotates together with the planet, which completely ignores the results of the analysis on the planet's surface. But they turned a deaf ear. They always impose the result analysing from the inertial system to the planet's surface.

 

I had to ask them to think about another question. Please see

api.ning.com/files/3-g0fF2x84RbBfYC8UIGakiS8f-yOI6UQyKWGQlgb9NJNhUPgZIouG6NcGyUGptucq1IiJHjztF2ogup7IVW1X884suQTr6J/file.JPG

 

Viewing from the rotating disk, is the person's journey about 100 meters or about 300 meters ?

Of course it is about 100 meters.

Here, why not impose the result viewing from the inertial system to the rotating disk?

 

Many people think that two beams of light can not return to point O at the same time because Sagnac effect.

They said Sagnac shows effect up in GPS satellites.

Yes, Sagnac effect shows up in GPS satellites. However, please note, the light speed relative to the GPS satellites is not C, but the C±V.

However, the light speed relative to the planet is not C±V, but C.

Sagnac effect exists because the light speed relative to the GPS satellites is C±V.

If the light speed relative to the GPS satellites is C, does Sagnac effect still exist?

 

My question is to reveal this.

 

I think that the two beams of light can return to point O at the same time.

If I am wrong, why the PRA's editors and reviewers do not point it out? Do they not understand physics ?

They attacked me in the past published papers, not this paper.

This shows I was right.

[swansont]

Once again you have ignored my question and changed the scenario. Since you will not stay on the topic of a specific discussion and answer question, I have no interest in discussing this further.

 

I think that the two beams of light can return to point O at the same time.

If I am wrong, why the PRA's editors and reviewers do not point it out? Do they not understand physics ?

They attacked me in the past published papers, not this paper.

This shows I was right.

 

You are the one who is wrong, and it is not the job of a journal to teach you physics. Reviewers will point out small mistakes that can be corrected. Blatant conceptual errors do not fall under this category. I don't see that they did anything wrong here.

[vuquta]

Many people can not understand why two beams of light can return to point O at the same time.

Michelson-Morley experiment let us know that the speeds of light in each direction are equal.

The question now is whether their journey is equal after gyrating a circuit they return to point O.

Many people think that they are not equal. Because of the rotation of the planet, the optical path of the light in counter-clockwise is less than that of the light in clockwise direction.

Although I remind that they analyzes this issue in the inertial system that never rotates together with the planet, which completely ignores the results of the analysis on the planet's surface. But they turned a deaf ear. They always impose the result analysing from the inertial system to the planet's surface.

 

I had to ask them to think about another question. Please see

api.ning.com/files/3-g0fF2x84RbBfYC8UIGakiS8f-yOI6UQyKWGQlgb9NJNhUPgZIouG6NcGyUGptucq1IiJHjztF2ogup7IVW1X884suQTr6J/file.JPG

 

Viewing from the rotating disk, is the person's journey about 100 meters or about 300 meters ?

Of course it is about 100 meters.

Here, why not impose the result viewing from the inertial system to the rotating disk?

 

Many people think that two beams of light can not return to point O at the same time because Sagnac effect.

They said Sagnac shows effect up in GPS satellites.

Yes, Sagnac effect shows up in GPS satellites. However, please note, the light speed relative to the GPS satellites is not C, but the C±V.

However, the light speed relative to the planet is not C±V, but C.

Sagnac effect exists because the light speed relative to the GPS satellites is C±V.

If the light speed relative to the GPS satellites is C, does Sagnac effect still exist?

 

My question is to reveal this.

 

I think that the two beams of light can return to point O at the same time.

If I am wrong, why the PRA's editors and reviewers do not point it out? Do they not understand physics ?

They attacked me in the past published papers, not this paper.

This shows I was right.

 

 

They said Sagnac shows effect up in GPS satellites.

Yes, Sagnac effect shows up in GPS satellites. However, please note, the light speed relative to the GPS satellites is not C, but the C±V.

However, the light speed relative to the planet is not C±V, but C.

Sagnac effect exists because the light speed relative to the GPS satellites is C±V.

 

Why does the satellite view C±V and what is v.

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Then you are doing the calculation incorrectly. Physically, why is the path a different length for clockwise vs counterclockwise travel? When the object is rotating, it's because the velocity vector of the light travel is parallel or antiparallel to the light travel — that's why you have c+v and c-v; it applies for the entire path. But for the effect of revolution this isn't true.

 

The revolution is a circular path just like Sagnac.

 

What logic causes this to be false? I do not understand this.

Edited April 16, 2010 by vuquta
Consecutive posts merged.

[swansont]

The revolution is a circular path just like Sagnac.

 

What logic causes this to be false? I do not understand this.

 

The light does not complete the circle along the path of the orbit. To first order, the path of the orbit is linear over that time.

[Xinwei Huang]

Once again you have ignored my question and changed the scenario. Since you will not stay on the topic of a specific discussion and answer question, I have no interest in discussing this further.

 

I am sorry. I am very busy recently, and my English is not good, I understand your question difficult.

I think I have said everything. But there are still many people can not recognize that relativity theory is wrong. I hope they can think deeply.

I believe that eventually people will think I am right.

[Mr Skeptic]

Did you think anyone would accept a theory like relativity if they didn't absolutely have to? If you have a better theory, one that has more accurate predictions (not make more sense), then maybe it could be replaced. Otherwise, don't bother.

[Xinwei Huang]

Did you think anyone would accept a theory like relativity if they didn't absolutely have to? If you have a better theory, one that has more accurate predictions (not make more sense), then maybe it could be replaced. Otherwise, don't bother.

 

Please see

http://www.scienceforums.net/forum/showthread.php?t=51328

[Mr Skeptic]

Right, but when you apply special relativity to a general relativity problem and get the wrong answer, that doesn't really prove that special relativity is wrong. It just means you made a mistake assuming an inertial frame where there is a non-inertial frame.